以下の関数を微分する問題です。 (1) $y = \tan^{-1} \sqrt{\frac{x-1}{2-x}}$ (2) $y = \log |\tan x|$ (4) $y = (\tan^{-1} 2x)^3$ (5) $y = \cos^{-1} x$ (7) $y = \tan^{-1} (\frac{1}{x})$

1. 問題の内容

以下の関数を微分する問題です。

(1) $y = \tan^{-1} \sqrt{\frac{x-1}{2-x}}$

(2) $y = \log |\tan x|$

(4) $y = (\tan^{-1} 2x)^3$

(5) $y = \cos^{-1} x$

(7) $y = \tan^{-1} (\frac{1}{x})$

2. 解き方の手順

(1) $y = \tan^{-1} \sqrt{\frac{x-1}{2-x}}$

合成関数の微分を行います。まず、$\tan^{-1} u$の微分は$\frac{1}{1+u^2}$です。次に、$\sqrt{v}$の微分は$\frac{1}{2\sqrt{v}}$です。最後に、$\frac{x-1}{2-x}$の微分は$\frac{(2-x) - (x-1)(-1)}{(2-x)^2} = \frac{2-x+x-1}{(2-x)^2} = \frac{1}{(2-x)^2}$です。したがって、

$\frac{dy}{dx} = \frac{1}{1 + \frac{x-1}{2-x}} \cdot \frac{1}{2\sqrt{\frac{x-1}{2-x}}} \cdot \frac{1}{(2-x)^2}$

$= \frac{2-x}{2-x+x-1} \cdot \frac{1}{2\sqrt{\frac{x-1}{2-x}}} \cdot \frac{1}{(2-x)^2}$

$= \frac{2-x}{1} \cdot \frac{1}{2\sqrt{\frac{x-1}{2-x}}} \cdot \frac{1}{(2-x)^2}$

$= \frac{1}{2(2-x)} \cdot \sqrt{\frac{2-x}{x-1}}$

$= \frac{1}{2(2-x)\sqrt{\frac{x-1}{2-x}}} = \frac{1}{2\sqrt{(2-x)^2\frac{x-1}{2-x}}}=\frac{1}{2\sqrt{(2-x)(x-1)}}$

(2) $y = \log |\tan x|$

$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) = \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x}$

(4) $y = (\tan^{-1} 2x)^3$

$\frac{dy}{dx} = 3(\tan^{-1} 2x)^2 \cdot \frac{1}{1 + (2x)^2} \cdot 2 = \frac{6(\tan^{-1} 2x)^2}{1+4x^2}$

(5) $y = \cos^{-1} x$

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$

(7) $y = \tan^{-1} (\frac{1}{x})$

$\frac{dy}{dx} = \frac{1}{1 + (\frac{1}{x})^2} \cdot (-\frac{1}{x^2}) = \frac{1}{1 + \frac{1}{x^2}} \cdot (-\frac{1}{x^2}) = \frac{x^2}{x^2+1} \cdot (-\frac{1}{x^2}) = -\frac{1}{x^2+1}$

3. 最終的な答え

(1) $\frac{1}{2\sqrt{(2-x)(x-1)}}$

(2) $\frac{2}{\sin 2x}$

(4) $\frac{6(\tan^{-1} 2x)^2}{1+4x^2}$

(5) $-\frac{1}{\sqrt{1-x^2}}$

(7) $-\frac{1}{x^2+1}$